High School

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If [tex]$f(4) = 246.4$[/tex] when [tex]$r = 0.04$[/tex] for the function [tex]$f(t) = P e^t$[/tex], then what is the approximate value of [tex]$P$[/tex]?

A. 50
B. 289
C. 1220
D. 210

Answer :

We are given the function

[tex]$$
f(t) = P e^{rt}
$$[/tex]

with the values

[tex]$$
f(4) = 246.4, \quad r = 0.04.
$$[/tex]

Since the function is defined as above, we substitute [tex]$t = 4$[/tex] into the function to get:

[tex]$$
246.4 = P e^{0.04 \times 4}.
$$[/tex]

First calculate the exponent:

[tex]$$
0.04 \times 4 = 0.16.
$$[/tex]

So the equation becomes:

[tex]$$
246.4 = P e^{0.16}.
$$[/tex]

To solve for [tex]$P$[/tex], we divide both sides of the equation by [tex]$e^{0.16}$[/tex]:

[tex]$$
P = \frac{246.4}{e^{0.16}}.
$$[/tex]

Evaluating the denominator, we have:

[tex]$$
e^{0.16} \approx 1.1735.
$$[/tex]

Thus,

[tex]$$
P \approx \frac{246.4}{1.1735} \approx 210.
$$[/tex]

Therefore, the approximate value of [tex]$P$[/tex] is [tex]$\boxed{210}$[/tex].

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